Name (print neatly your full name)________________________________________________________________
Solutions to Midterm I written portion (100 points worth 75% of Midterm I grade) – Solar Yellow version
1. (15 points as broken down below) Agan Interior Design provides home and office decorating assistance to its customers. In normal
operation, an average of 2.5 customers arrive each hour. One design consultant is available to answer customer questions and make
product recommendations. The consultant averages 10 minutes with each customer.
a. (5 points) Compute the operating characteristics of the customer waiting line, assuming Poisson arrivals and
exponential service times.
b. (5 points) Service goals dictate that an arriving customer should not wait for service more than an average of 5
minutes. Is this goal being met?
c. (5 points) If the consultant can reduce the average time spent per customer to 8 minutes, what is the mean service
rate? Will the service goal be met?
a. ， = 2.5 ( = 60/10 = 6 customers per hour
LqW==0.1190hours (7.14 minutes)q，
b. No; W = 7.14 minutes. Firm should increase the mean service rate (µ) for the consultant or hire a second consultant. q
c. ( = 60/8 = 7.5 customers per hour
Lq==0.0667hours (4 minutes)Wq，
The service goal is being met.
2. (15 points) A fast food franchise is considering operating a drive-up window food-service operation. Assume that customer arrivals follow a Poisson probability distribution, with a mean arrival rate of 24 cars per hour, and that service times follow an exponential
probability distribution. Arriving customers place orders at an intercom station at the back of the parking lot and then drive to the
service window to pay for and receive their orders. The following three service alternatives are being considered.
： A single-channel operation in which one employee fills the order and takes the money from the customer. The average
service time for this alternative is 2 minutes.
： A single-channel operation in which one employee fills the order while a second employee takes the money from the
customer. The average service time for this alternative is 1.25 minutes.
： A two-channel operation with two service windows and two employees. The employee stationed at each window fills
the order and takes the money for customers arriving at the window. The average service time for this alternative is 2
minutes for each channel.
All of the following must be carefully analyzed to aid in the economic analysis:
o Customer waiting time is valued at $25 per hour to reflect that waiting time is costly to the fast food business.
o The cost of each employee is $6.50 per hour.
o To account for equipment and space, an additional cost of $20 per hour is attributable to each channel.
Based on the steady-state queueing theory that we have studied in ISE 521, what is the lowest-cost of the 3 designs for the fast
， = 24
System A System B System C
Characteristic (k = 1, µ = 30) (k = 1, µ = 48) (k = 2, µ = 30)
a. P 0.2000 0.5000 0.4286 0
b. L 3.2000 0.5000 0.1524 q
d. W 0.1333 0.0200 0.0063 q
d. W 0.1667 0.0417 0.0397
e. L 4.0000 1.0000 0.9524
f. P 0.8000 0.5000 0.2286 w
System C provides the best service, but what about cost?
Service Cost per Channel
System A: 6.50 + 20.00 = $26.50/hour
System B: 2(6.50) + 20.00 = $33.00/hour
System C: 6.50 + 20.00 = $26.50/hour
Total Cost = cL + ck ws
System A: 25(4) + 26.50(1) = $126.50
System B: 25(1) + 33.00(1) = $ 58.00
System C: 25(0.9524) + 26.50(2) = $ 76.81
System B is the most economical.
nd3. (20 points) Pam’s Pub & Advice Emporium (PPAE) operates out of a 2 floor room in Baker Systems Engineering. Customers arrive in a Poisson fashion at a rate of 24/hour. Service times are exponential with a
mean of 2.0 minutes. The PPAE only allows 4 customers to be in the room regardless of the number of thservers waiting on the clientele. When more than 4 people are in the system, the extra customers (i.e., the 5, th6,… currently in the system) must wait in the hallway. The ISE safety commission gets nervous when people ndstand outside the PPAE waiting. Currently the PPAE has one server (Pam) but is considering adding a 2
server (Pam2) if it can demonstrate that the probability of customers waiting in the hallway can be reduced by at
least a factor of 10 (or in other words be < 10% of the probability of people waiting in the hallway with 1 server). What do you recommend and why? You must have sound numerical support to specifically address
the probability of people waiting in the hallway.
The initial system with 1 server is M/M/1 and it needs to be compared to a M/M/2 system with two servers.
，！24,(！30.Working in hours (you can do it in minutes if you so desire), we find that For the M/M/1 system,
the P( x > 4) = 1 – P(x < 4) = 1 – (.20 + .16 + .128 + .102 + .082) = 1- 0.672 = 0.328.
Doing the same math for the M/M/2 (2 servers) we get P( x > 4) = 1 – (.429 + .343 +.137 +.055 +.022) = 1 -
.986 = 0.014. The M/M/2 system P( x > 4) < 10% [M/M/1 P(x > 4)] = .10 * 0.328 = 0.0328; therefore, the M/M/2 is the preferred system based on the criteria posed in the problem.
4. (20 points as broken down below) George was running the annual Hug-an-IE fund raiser. While the huggers trade places with other during the hugathon, there are always 4 on duty. It is assumed that a single line of hugees (those desiring to hug an IE) is used and each hugee is pleased to hug whoever is the next available hugger regardless of the sex of both parties. The arrival rate of hugees follows a Poisson distribution with an average of one per minute. Each of the four huggers provides exponential service and averages 30 hugs per hour. Pay close attention to the proper use of units and also to the meaning of the different parameters involved with queueing.
a. (5 points) What is the probability that all 4 huggers are currently in use and there is no one in the
Answer: This is a M/M/4 system where the huggers are the servers. The arrival rate is
，(！！6030..andservicerateperkegThesearebothperhour With all 4 huggers in use and no one in the
queue, we want P from the P formula. Since k = 4 and n = 4, we are going to use the formula for n < k. First 4n
we have to look up P in table 14.4 which turns out to be = 0.1304. Then do the following: 0
b. (10 points) What is the probability that all 4 huggers are currently in use?
Answer: M/M/4 still with same parameters, but now we want
Or just compute P using formula (15.16) and get P = (1/24)*(16)*((4*30)/(4*30-60))*0.1304 = 0.174 ww
c. (5 points) What is the probability that a single patron’s will get a hug between ? and 3 minutes?
Answer: This is just an application of the exponential distribution. You must be careful to put in the exponential parameter value correctly which depends on what form of the exponential distribution is used. Using the version given in the queueing handout where the mean is a rate, we have the following making sure you use compatible time units. Hours are used below.
5. (25 points) Arena problem: A variation on the ISE Hugathon. In reality the real Hugathon cannot be modeled via closed form steady-state queueing theory as accurately as desired because people have preferences on who they get to hug. Assume that 12 huggers are working for our simulation purposes with ? being of each sex. (Or if it makes it easier for you, use 6 of one sex and half a dozen of the other). Assume the incoming population is 60% male. Let the huggers be known as Women1, Women2, …, Women6, Man1, Man2, …, Man6. Customers arrive at the same rate as in the earlier queueing problem (keep in mind the relationship between the Poisson and exponential – it is in the handout in case you don’t remember it).
80% of the male huggees arriving want a female hugger and are equally pleased with any of the 6 women available. 20% of the male huggees don’t care who they get and are equally pleased with any of the 12 huggers.
95% of the women huggees prefer men only in the following order: Man1, Man3, Man5, Man2, Man4, Man6 but will take the next available man hugger even if their initial choices are busy. 5% of the women don’t care who they hug and are equally pleased with any of the 12 huggers.
Using Resource Sets in any & all Process or Seize modules to be efficient and assuming that the hugging distribution times are the same as given in the earlier queueing problem, develop the Arena simulation. Show all the necessary modules as boxes (or shapes) linked together as needed. Label each box (or shape) so that I know whether it is a Create, Process, Decide, or whatever. List only the things you need to change from the
defaults to get the model to run. Show the insides of any spreadsheet-like icon that is needed. Do not worry about things like how long the simulation run time is.
6. (5 points) Arena Cycle time question: how would one capture it separately in the output when there are two different part types? There are various ways to do this. Fully describe one of them. Do not cover more than just one method.
Answer: One way is to have different Entity Types and click on that you want entity statistics in Run, Setup on the Project Parameters tab as seen below:
Another way is to create an attribute (e.g., ArrivalTime) and set it = TNOW in an Assign module right after the entities are created in a Create module. Then you could use a Record module to capture the Time Interval statistic just prior to a Dispose Module. To capture them separately for each entity type, you could record them into a set assuming you have created an index variable to that set. Or you could just direct the different entity types to different record modules and label the outputs accordingly.